Image Processing Reference
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Therefore,
4z þ
6
:
4
H(z)
¼
z 2
þ
0
:
5z þ
0
:
6
4.8 SOLUTION OF LTI DISCRETE-TIME STATE EQUATIONS
Consider a discrete LTI system described by the state-space equation
x ( k þ
1
) ¼ Ax ( k ) þ Bu ( k )
(
4
:
100
)
and the output equation
y ( k ) ¼ Cx ( k ) þ Du ( k )
(
:
)
4
101
To obtain the output of this system for a given input and an initial state, we
first solve
the state equation (Equation 4.100) and then the solution is substituted into the
algebraic equation (Equation 4.101) in order to
. The solution to
state equations has two parts, a homogeneous solution and a particular solution. We
nd the output y ( k )
first consider the homogeneous solution.
4.8.1 S OLUTION OF H OMOGENEOUS S TATE E QUATION
The homogeneous solution is the solution of state equations to an arbitrary initial
condition with zero input. The homogeneous state equation is given by
x ( k þ
1
) ¼ Ax ( k )
(
4
:
102
)
Assuming that the initial state is x (
0
)
, we have
x (
1
) ¼ Ax (
0
)
) ¼ A 2 x (
x (
2
) ¼ Ax (
1
) ¼ A Ax (
0
0
)
) ¼ A A 2 x (
) ¼ A 3 x (
x (
3
) ¼ Ax (
2
0
0
)
(
4
:
103
)
.
x ( k ) ¼ Ax ( k
) ¼ A A k 1 x (
) ¼ A k x (
1
0
0
)
Therefore, in matrix form, the solution to the homogeneous equation is
x ( k ) ¼ A k x (
0
)
(
4
:
104
)
Therefore, the homogeneous part of the solution is x ( k ) ¼ A k x (
0
)
. The matrix expo-
nential A k is called state-transition matrix and is denoted by
w( k )
. The state-transition
matrix
w( k )
is an N N matrix and is the solution to the homogeneous equation
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